| Exam Board | OCR |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2014 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating the Binomial to the Poisson distribution |
| Type | Explain why approximation inappropriate |
| Difficulty | Standard +0.3 This question tests understanding of when Poisson and normal approximations to binomial are valid, requiring recall of conditions (n large, p small for Poisson; np and nq both >5 for normal) and the insight to consider the complementary event. While it requires conceptual understanding rather than pure calculation, the conditions are standard bookwork and the 'consider non-attendance' hint makes part (ii) straightforward, placing it slightly easier than average. |
| Spec | 2.04d Normal approximation to binomial5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(np = 147 > 5\) so not Poisson | M1 | Consider any two conditions out of \(np\), \(nq\) (allow \(npq\)), size of \(n\), size of \(p\) |
| 147 stated, or "\(p\) not small", no wrong conditions for Poisson seen | A1 | |
| \(nq = 3 < 5\) so not normal | A1 | 3 [not *just* 2.94] stated, or "\(p\) not close to \(\frac{1}{2}\)"; no wrong conditions for normal seen (apart from \(npq\)). *If spurious extra reasons seen ("not independent" etc), max 2/3* |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(A \sim B(150, 0.98)\) so \(150 - A \sim B(150, 0.02)\) | M1 | Clearly consider complement with \(p = 0.02\) |
| \(\approx Po(3)\) | A1 | \(Po(3)\) stated or implied |
| \(P(A < 146) = P(150 - A > 4) = 1 - 0.8153\) | M1 | \(1 - Po(3)\) probability, e.g. 0.3528 or 0.0839 |
| \(= \mathbf{0.1847}\) | A1 | 0.185 or better. [Exact binomial (0.1830): 0/4. \(N(3, 2.94)\): M1A0M0A0] |
| [4] |
## Question 2:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $np = 147 > 5$ so not Poisson | M1 | Consider any two conditions out of $np$, $nq$ (allow $npq$), size of $n$, size of $p$ |
| 147 stated, or "$p$ not small", no wrong conditions for Poisson seen | A1 | |
| $nq = 3 < 5$ so not normal | A1 | 3 [not *just* 2.94] stated, or "$p$ not close to $\frac{1}{2}$"; no wrong conditions for normal seen (apart from $npq$). *If spurious extra reasons seen ("not independent" etc), max 2/3* |
| | **[3]** | |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $A \sim B(150, 0.98)$ so $150 - A \sim B(150, 0.02)$ | M1 | Clearly consider complement with $p = 0.02$ |
| $\approx Po(3)$ | A1 | $Po(3)$ stated or implied |
| $P(A < 146) = P(150 - A > 4) = 1 - 0.8153$ | M1 | $1 - Po(3)$ probability, e.g. 0.3528 or 0.0839 |
| $= \mathbf{0.1847}$ | A1 | 0.185 or better. [Exact binomial (0.1830): 0/4. $N(3, 2.94)$: M1A0M0A0] |
| | **[4]** | |
---2 The events organiser of a school sends out invitations to $\mathbf { 1 5 0 }$ people to attend its prize day. From past experience the organiser knows that the number of those who will come to the prize day can be modelled by the distribution $\mathbf { B } ( \mathbf { 1 5 0 } , \mathbf { 0 . 9 8 } )$.\\[0pt]
(i) Explain why this distribution cannot be well approximated by either a normal or a Poisson distribution. [3]\\[0pt]
(ii) By considering the number of those who do not attend, use a suitable approximation to find the probability that fewer than 146 people attend. [4]
\hfill \mbox{\textit{OCR S2 2014 Q2 [7]}}